You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.

Section 7-1 : Proof of Various Limit Properties

In this section we are going to prove some of the basic properties and facts about limits that we saw in the Limits chapter. Before proceeding with any of the proofs we should note that many of the proofs use the precise definition of the limit and it is assumed that not only have you read that section but that you have a fairly good feel for doing that kind of proof. If you’re not very comfortable using the definition of the limit to prove limits you’ll find many of the proofs in this section difficult to follow.

The proofs that we’ll be doing here will not be quite as detailed as those in the precise definition of the limit section. The “proofs” that we did in that section first did some work to get a guess for the \(\delta \)and then we verified the guess. The reality is that often the work to get the guess is not shown and the guess for \(\delta \)is just written down and then verified. For the proofs in this section where a \(\delta \) is actually chosen we’ll do it that way. To make matters worse, in some of the proofs in this section work very differently from those that were in the limit definition section.

So, with that out of the way, let’s get to the proofs.

Limit Properties

In the Limit Properties section we gave several properties of limits. We’ll prove most of them here. First, let’s recall the properties here so we have them in front of us. We’ll also be making a small change to the notation to make the proofs go a little easier. Here are the properties for reference purposes.

Note that we added values (\(K\), \(L\), etc.) to each of the limits to make the proofs much easier. In these proofs we’ll be using the fact that we know \(\mathop {\lim }\limits_{x \to a} f\left( x \right) = K\) and \(\mathop {\lim }\limits_{x \to a} g\left( x \right) = L\) we’ll use the definition of the limit to make a statement about \(\left| {f\left( x \right) - K} \right|\) and \(\left| {g\left( x \right) - L} \right|\) which will then be used to prove what we actually want to prove. When you see these statements do not worry too much about why we chose them as we did. The reason will become apparent once the proof is done.

Also, we’re not going to be doing the proofs in the order they are written above. Some of the proofs will be easier if we’ve got some of the others proved first.

Proof of 7

This is a very simple proof. To make the notation a little clearer let’s define the function \(f\left( x \right) = c\) then what we’re being asked to prove is that \(\mathop {\lim }\limits_{x \to a} f\left( x \right) = c\). So let’s do that.

Let \(\varepsilon > 0\) and we need to show that we can find a \(\delta > 0\) so that

The left inequality is trivially satisfied for any \(x\) however because we defined \(f\left( x \right) = c\). So simply choose \(\delta > 0\) to be any number you want (you generally can’t do this with these proofs). Then,

Proof of 1

There are several ways to prove this part. If you accept 3 And 7 then all you need to do is let \(g\left( x \right) = c\) and then this is a direct result of 3 and 7. However, we’d like to do a more rigorous mathematical proof. So here is that proof.

In the third step we used the fact that, by our choice of \(\delta \), we also have \(0 < \left| {x - a} \right| < {\delta _{\,1}}\) and \(0 < \left| {x - a} \right| < {\delta _{\,2}}\) and so we can use the initial statements in our proof.

Next, we need to prove \(\mathop {\lim }\limits_{x \to a} \left[ {f\left( x \right) - g\left( x \right)} \right] = K - L\). We could do a similar proof as we did above for the sum of two functions. However, we might as well take advantage of the fact that we’ve proven this for a sum and that we’ve also proven 1.

Proof of 3

This one is a little tricky. First, let’s note that because \(\mathop {\lim }\limits_{x \to a} f\left( x \right) = K\) and \(\mathop {\lim }\limits_{x \to a} g\left( x \right) = L\) we can use 2 and 7 to prove the following two limits.

Fairly simple proof really, once you see all the steps that you have to take before you even start. The second step made multiple uses of property 2. In the third step we used the limit we initially proved. In the fourth step we used properties 1 and 7. Finally, we just did some simplification.

In the second step we could remove the absolute value bars by adding in the negative because we know that \(f\left( x \right) > 0\) and can safely assume that \(g\left( x \right) < 0\) (as noted above).

So, provided \(c > 0\) we’ve proven that \(\mathop {\lim }\limits_{x \to \infty } \frac{c}{{{x^r}}} = 0\). Note that the main difference here is that we need to take the absolute value first to deal with the minus sign. Because both sides are negative we know that when we take the absolute value of both sides the direction of the inequality will have to switch as well.

Case 2, Case 3 : As noted above these are identical to the proof of the corresponding cases in the first proof and so are omitted here.

We’re going to prove this in an identical fashion to the problems that we worked in this section involving polynomials. We’ll first factor out \({a_n}{x^n}\) from the polynomial and then make a giant use of Fact 1 (which we just proved above) and the basic properties of limits.

Now, clearly the limit of the second term is one and the limit of the first term will be either \(\infty \) or \( - \infty \) depending upon the sign of \({a_n}\). Therefore by the Facts from the Infinite Limits section we can see that the limit of the whole polynomial will be the same as the limit of the first term or,

Let’s start with the fact that \(f\left( x \right)\) is continuous at \(x = b\). Recall that this means that \(\mathop {\lim }\limits_{x \to b} f\left( x \right) = f\left( b \right)\) and so there must be a \({\delta _{\,1}} > 0\) so that,

But, we also know that if \(0 < \left| {x - b} \right| < {\delta _{\,1}}\) then we must also have \(\left| {f\left( x \right) - f\left( b \right)} \right| < \varepsilon \). What this is telling us is that if a number is within a distance of \({\delta _{\,1}}\) of \(b\) then we can plug that number into \(f\left( x \right)\) and we’ll be within a distance of \(\varepsilon \) of \(f\left( b \right)\).

So, \[\left| {g\left( x \right) - b} \right| < {\delta _{\,1}}\] is telling us that \(g\left( x \right)\) is within a distance of \({\delta _{\,1}}\) of \(b\) and so if we plug it into \(f\left( x \right)\) we’ll get,